Question: Suppose that $f(x)=4x+5$. What is $f^{-1}(f^{-1}(9))$?
Inputting $f^{-1}(x)$ into $f$, we have $f(f^{-1}(x)) =4f^{-1}(x) + 5$, so $x = 4f^{-1}(x) + 5$.  Solving this equation for $f^{-1}(x)$, we get that $f^{-1}(x) = \frac{x-5}{4}$. Thus, we have  \begin{align*}
f^{-1}(f^{-1}(9)) & = f^{-1}\left(\frac{9-5}{4}\right) \\
& = f^{-1}(1) \\
& = \frac{1-5}{4} \\
& = \boxed{-1}.
\end{align*}